Nitrogen gas (N2) can be prepared by passing gaseous ammonia (NH3) over solid copper (ll) oxide (CuO)?

Question

Nitrogen gas (N2) can be prepared by passing gaseous ammonia (NH3) over solid copper (ll) oxide (CuO) at high temperatures. The other products of the reaction are solid copper (Cu) and water vapor (H2O)
a. If 18.1 g of NH3 is reacted with 90.4 g of CuO, which is the limiting reactant?
b. How many grams of N2 will be formed?
plz help me !!!!!

Answer 1

The balanced equation is :
2NH3 +3 CuO >> N2 + 3 Cu +3 H2O

a.
molar mass NH3 = 17 g/mol
18.1 g / 17 = 1.06 mole NH3

Molar mass CuO = 79.55 g/mol
90.4 g / 79.55 = 1.14 mole CuO
The ratio between NH3 and CuO is 2 : 3 so CuO is the limiting reactant

The ratio between CuO ( limiting reactant ) and N2 is 3 : 1
3 : 1 = 1.14 : x
x = moles N2 = 0.38
Molar mass N2 = 28 g/mol
0.38 mol x 28 g/mol = 10.6 g of N2 produced

Answer 2

First balance the equation:

2NH3 + 3CuO --> 3Cu + 3H2O + N2

Then look up the atomic weights of each of the substances and work out the molecular weights, e.g. NH3 = 14+1+1+1=17. Write these under the various substances, remembering that for NH3, for example, you have 2 molecules of gas so you use 34.

If I've calculated correctly, the numbers should be something like:

34 + 238.8 --> 190.8 + 54 + 28

The limiting reactant is CuO because you have 90.4g of it which is 90.4/238.8 = 0.379 of what your formula suggests, whereas for NH3 you have 18.1/34 = 0.532 (therefore surplus compared with CuO.)

The amount of N2 formed will therefore be 0.379 * 28 = 10.6g.

Hope I've done the sums right - the principle is correct!